Comments for Mathematica Cookbook

Comment on Challenge Answer by Sal Mangano

Sal Mangano — Sat, 14 Apr 2012 02:31:41 +0000

Fixed the formatting. Thanks for pointing that out.

If the Sort is a stable sort then this would be the case because this was the order prior to sorting. In any case the length of the words themselves would not have any bearing on the order because SortBy is only considering the size of each family.

Comment on Challenge Answer by Dan

Dan — Sat, 14 Apr 2012 02:16:48 +0000

Elegant solution! (It would be visually neater if the “DictionaryLookup” were on a new line so the whole code example could be viewed at the same time.)

The next-to-last anagram set also had eight elements. But in your sorting and Simon’s, the 5-letter words was deemed longest. What specifically accounts for that?

Thanks for a nice challenge with a neat solution!
Dan

Comment on Challenge by Challenge Answer | Mathematica Cookbook

Challenge Answer | Mathematica Cookbook — Sat, 14 Apr 2012 01:55:09 +0000

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Comment on Challenge by Sal Mangano

Sal Mangano — Fri, 13 Apr 2012 14:16:30 +0000

My motivation is that I was doing research for a iOS project where I have to write an anagram generator in Objective C.

Comment on Challenge by Simon Tyler

Simon Tyler — Fri, 13 Apr 2012 14:08:33 +0000

@Sal, the last line of my comment said the same thing worked starting with the whole dictionary… I just liked the Repeated[_, {5}] pattern, it’s something I don’t get to use often.

What was the motivation for you writing the post? Was there some reason you were playing around with the dictionary data? Or was it just for fun?

Comment on Challenge by Dan

Dan — Fri, 13 Apr 2012 12:36:37 +0000

By “Yes,” of course, I meant “no” to my question. But it was interesting that WordData and DictionaryLookup give different answers:
DL: {“least”, “slate”, “Stael”, “stale”, “steal”, “tales”, “teals”, “Tesla”}

WD: {“Lates”, “least”, “slate”, “Stael”, “stale”, “steal”, “stela”, “tesla”, “Tesla”}

Comment on Challenge by Dan

Dan — Fri, 13 Apr 2012 12:29:43 +0000

Yes, run Simon’s code on DictionaryLookup[] instead.

Comment on Challenge by Dan

Dan — Fri, 13 Apr 2012 12:26:07 +0000

It’s necessary to use _some_ information about the output set. Is it that it’s a set of 8 anagrams? That is, for each word in the dictionary generate the list of its anagrams, and choose the entry with the longest length.

Comment on Challenge by Sal Mangano

Sal Mangano — Fri, 13 Apr 2012 12:02:05 +0000

@Simon That is close but your solution starts out knowing the answer has words of length 5 and mine does not.

Comment on Challenge by Sal Mangano

Sal Mangano — Fri, 13 Apr 2012 11:57:56 +0000

@Dan You can suggest an answer and then maybe some one will code it!